How to Calculate Time for a Mass on a Spring to Reach Equilibrium Again?

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Homework Help Overview

The problem involves a mass-spring system where a mass of 1.64 kg stretches a spring and is then released from an additional stretch. The goal is to determine the time it takes for the mass to return to the new equilibrium position.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the spring constant and the time to reach equilibrium using various physics equations related to forces and motion.
  • Some participants question the assumptions made about the nature of the motion, suggesting that it is not with constant acceleration.
  • Others suggest using the period of vibration to find the time required for the mass to return to equilibrium.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been provided regarding the period of vibration and the relationship between the period and the time to reach equilibrium.

Contextual Notes

There seems to be confusion regarding the application of formulas and the nature of the motion involved in the mass-spring system. The original poster expresses uncertainty about their calculations and assumptions.

ninjagowoowoo
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Q:
A mass of 1.64 kg stretches a vertical spring 0.300 m. If the spring is stretched an additional 0.123 m and released, how long does it take to reach the (new) equilibrium position again?

What I tried:
Found k using F=kx (mg=kx)
using that k, I calculated the force in the extended spring (F=k(0.123))
using that force, I calculated the acceleration of the mass (F=ma)
finally, using x=(1/2)at^2 I calculated t.

I must be missing something obvious b/c this doesn't seem like a difficult problem... :confused:
 
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Nope,it's incorrect.The motion is not with constant acceleration.U need to find the period of vibration.

Daniel.
 
Yeah that's what I thought...

So I calculated k in the first step above.
Then I used omega = sqrt(k/m) to get omega.
Then I used f = omega/2pi
Then period = 1/f and it's still wrong. Any other mistakes I made?

Thanks for the help :smile:
 
Yes,the time required is only 1/4-th of the period...Can u see why?

Daniel.
 
Wow hey thanks a lot. I guess brain just isn't working today. :smile:
 

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